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Foods that are high in energy will boost the metabolism while if you eat foods that are low in energy, your metabolism will slow down. This is because you have a certain set point for your fat reserves, the body strives to maintain this on the long run. Count Iblis (talk) 02:29, 21 September 2012 (UTC)[reply]
I don't think that's correct -- can you show references to back it up? My understanding is that for a person in good health who eats an adequate amount of food, without eating extreme amounts of one specific type such as protein or sugar, the type of food does not have a major effect on metabolic rate. This leaves out foods and beverages that function as drugs, such as coffee, alcohol, ephedra, etc. Looie496 (talk) 04:48, 21 September 2012 (UTC)[reply]
I agree with Looie496 foods, per say, do not affect the metabolic rate of the body to produce more energy. If there were foods that did that then we would have the solution to several medical problems, indeed if there were foods that slowed the human metabolism just think of the advantages in treating thyrotoxicosis for example. Obesity would no longer be a problem if we had some sort of foodstuff that could be used to burn off the fat. In spite of misleading adverts these foods don't exist. And the 'sugar-rush' is psychological in basis in my opinion, (sorry, that's OR). Richard Avery (talk) 07:41, 21 September 2012 (UTC)[reply]
I believe there are foods which increase the metabolic rate, such as those which contain stimulants like caffeine. However, this is a small effect, and people can always just eat more calories to compensate. Also, increasing the metabolic rate isn't really what you want to do anyway, since that means more oxidation of the cells, causing more damage. A lower metabolic rate, with fewer calories consumed, is the better way to go for your long-term health. StuRat (talk) 08:12, 21 September 2012 (UTC)[reply]
The total amount of calories consumed will influence the metabolic rate so that the weight is kept stable on the long term, see e.g. here. While the program focusses on thin people who know from experience that their weight is independent on how much they eat, this is to some degree true for everyone.
Suppose that you reduce your calorie intake by 100 Kcal/day. Then assuming there is no relation between calorie intake and metabolic rate, this should lead to 1 kg body weight reduction every 80 days, so that's more than 45 kg weight loss in ten years. Yet, we all know that you won't lose 45 kg body weight by eating a sandwich a day less. Therefore, this proves that the calorie intake influences the metabolic rate to keep your weight on the long term stable. Of course, what really matters here is the balance between energy intake and energy use. You could also consider eating the same amount and exercising a bit more every day. Then you won't keep on losing weight, your weight will stabilize at some level. So, a new equilibrium between energy intake and energy expended will be reached.
Only if there is a deficit on the energy balance that is beyond what the body can compensate for, will you face starvation. But then the opposite is also true, eating more does not lead to long term weight gain, unless you exceed a certain threshold (which is different for different people). Count Iblis (talk) 16:39, 21 September 2012 (UTC)[reply]
I don't think that's quite right:
1) Eating less does not automatically decrease your BSL, unless you eat so little as to activate your body's starvation response.
2) Eating more does not automatically increase your BML, unless your body was in starvation mode to begin with.
3) I agree that skipping one small item of food each day does not automatically lead to weight loss. However, I disagree as to the mechanism. When you skip something, this leaves you hungrier, so you eat and drink a bit more of everything else. If your diet was precisely controlled to prevent you from eating and drinking more, then you would, indeed, lose weight by cutting one small item from a diet which otherwise would lead you to maintain a stable weight. StuRat (talk) 16:55, 21 September 2012 (UTC)[reply]
The article on PCS mentions, but does not explain why, a severe onset of PCS may cause blurred vision in conjunction with the pain. Is this a pain response of some kind? Is there a biological rationale behind it or is it about some manner of overwhelming sensation? I've attempted to google it and to look through references for the PCS article, but little came of that. If someone could elaborate I would sure appreciate it, so thank you in advance for any help!
Can't find much about it, the condition is not mentioned in the Merck manual. I'm not sure if an answer is available; in my experience, conditions considered not dangerous and requiring no treatment receive little attention. The end of the article's history section suggests that your question may remain unanswered: These constitute the literature available on PCS. A search of Pubmed shows only six publications on the topic, five of them quoted in the wikipedia article, and two mentioning the syndrome. So if it's a symptom of PCS itself, you're unlikely to find more info. Your idea of it being a pain response could be correct, trouble is finding a source with google, considering all the webpages on eye diseases, migraine, myofascial pain syndrome, fibromyalgia, hypertension, postpartum changes, Polymyalgia rheumatica, drug side effects and so on. I'm having no luck so far. Ssscienccce (talk) 12:12, 21 September 2012 (UTC)[reply]
While it may not be dangerous in itself, the potential confusion with heart attacks could lead people with heart attacks to go untreated, at great risk to themselves. StuRat (talk) 16:59, 21 September 2012 (UTC)[reply]
Google Scholar gives better results, but not good enough. The syndrome is extremely common, but little studied because of its transience and lack of health effects. [1] suggests it should really be called "thoracic catch" because it isn't limited to the area over the heart. I will add (not as an answer but merely as an addition to the question) that when I was younger I remember encountering something like this (but never formally diagnosed, and not with any blurring of vision) now and then - my subjective impression was that two opposing surfaces somewhere in the lung were sticking together, and my reaction to it was to very slowly inhale despite the pain in order to force them apart, which seemed always to work. Wnt (talk) 15:33, 22 September 2012 (UTC)[reply]
I have a baby(4 in.)snapping turtle pet. He or she, I think, is trying to hibernate by going inside a conk shell. When do snapping turtles hibernate(month)? Will it be ok to keep him in the 30 gallon aquarium? I will turn him lose if necessary to save his life. Please advise. Thanks! — Preceding unsigned comment added by 66.190.198.190 (talk) 09:28, 21 September 2012 (UTC)[reply]
I'm at work, so I can't look up reference right now, but I know there is plenty of information if you search for things like "pet turtle hibernation". A 30 gallon tank will be fine for one turtle that size, assuming you have it set up well. The turtle will probably spend a lot of time buried or semi-buried during the winter months. I'm basing this on my parents' turtles - red-eared sliders raised from eggs when my dad dug into a nest while modifying our pond's bank. They are currently using a large tank (probably 40 gallons) that is filled with water. There is a soil-filled plastic tub that takes up the upper-left corner of the aquarium, with a wooden ramp they can use to climb up to it. The tub is large enough to have a basking area with a heat lamp and a cooler, damper area. They designed it based on what they found researching aquarium designs for turtles. If I remember, I'll look for some references for you tonight when I get home. 209.131.76.183 (talk) 12:19, 21 September 2012 (UTC)[reply]
Please note that people on the Refdesk are not giving you advice. Medical advice, which I assume includes veterinary, is especially not being given here. Note that the site disclaimer indicates that we could be deliberately confusing you for the fun of it and you have no recourse - at best, you're getting someone who has no verifiable credentials giving you a guess. I'd strongly suggest you rephrase to ask for information, not advice, such as where you can find informative sites about raising a baby snapping turtle. (And if I were going to give any advice, it would be that keeping a pet snapping turtle sounds like a really bad idea no matter how you do it ...) Wnt (talk) 21:30, 21 September 2012 (UTC)[reply]
Wonderful. Now i will try to explain you in few (busy at present in some other work) days how ancient Indian's were resistant for psychologically suffering. Only Ancient Indian philosophy have the answer not only for psychological but also for other sufferings. — Preceding unsigned comment added by Drcsjain (talk • contribs) 11:03, 21 September 2012 (UTC)[reply]
Actually we'd be grateful if you didn't try to explain anything to us. This is a Reference desk and we do the explaining. It is not a forum for debate, nor is it the place to evangelise. --TammyMoet (talk) 12:14, 21 September 2012 (UTC)[reply]
No. Mason jars are just annealedSoda-lime glass. So they are reasonable durable. To use Toughened glass would just make them cost more without any providing any advantage. When you drop one (or in my case - a whole tray-full) they shatter just like ordinary glass. They are just inexpensive containers that can be used many times... The original Kilner jars are the best. Their quality far exceeds anything made in the US and else where. You can pass them on to your grand-children. A most wise investment!--Aspro (talk) 15:34, 21 September 2012 (UTC)[reply]
Maybe your question comes from the fact that they are heat sterilized in boiling water or steam, and cookware is sometimes made with tempered glass (like US-made Pyrex). It's not the temperature that causes glass to break but temperature differences in the glass. It's safe to heat a jar by submerging in water and heating the water, but putting a cold jar into hot water can break it, especially if only partly submerged because of the stress between the hot and cold part (due to thermal expansion). Since the jars are only heated on those occasions (unlike oven cookware), and it can be safely done by following proper procedure, there's no point in using much more expensive glassware for the job. Ssscienccce (talk) 17:26, 21 September 2012 (UTC)[reply]
During the First World War it was found that quick lime decomposed all sorts of nasty gases and stuff. Since then, many companies have tried to patten this technology only to find their pattens are invalid due to prior art. This seems to be one of the latest. Look for 'Chlorofluorocarbon' (CFC) in this docucument: [2] Every few years you will see a press releases, stating that such an' such company has just made a breakthrough in destroying chlorinates - Blah! It all been done before. --Aspro (talk) 23:38, 21 September 2012 (UTC)[reply]
If a photon is massless then the ratio of photon to universe is always the same despite the fact that the universe is expanding, right? But no so for massive particles?165.212.189.187 (talk) 14:58, 21 September 2012 (UTC)[reply]
? Can you rephrase that? I can see what your getting at (but a photon's mass at rest is 'one' so when we consider the 'Universe' and everything.....). Whoops, maybe I should rephrase that too--Aspro (talk) 15:48, 21 September 2012 (UTC)[reply]
A photon is not a number, and the universe is not a number, so they don't have a ratio. I agree that the question needs to be clarified. Looie496 (talk) 16:10, 21 September 2012 (UTC)[reply]
How about this: If a photon is massless then the ratio of photon volume to universe volume is always the same despite the fact that the universe is expanding, right? But no so for massive particles?165.212.189.187 (talk) 16:16, 21 September 2012 (UTC)[reply]
The Universe's "volume" is an unknown unknown. That is, we don't have any consensus model for the shape of the universe, so we can't even say for certain whether the universe is "open" or "closed", "bounded" or "boundless". It's a nonsensical question. It's kinda the physics equivalent of asking "Do colorless green ideas really sleep furiously", it has all of the bits that seem to make up a physics question, but it's fairly nonsensical given what we know about what the "universe" is and what a "photon" is. --Jayron3216:23, 21 September 2012 (UTC)[reply]
The universe's volume isn't an unknown unknown. You don't know what the unknown unknowns are — that's what's the first unknown signifies. Unknown unknowns are the things you don't even realize you don't know (whoa, dude, deep). The volume of the universe is a known unknown — something we know that we don't know. It's not a nonsense question to ask about the volume of the universe, just because we don't have at the moment a good consensus on it. It's not the same class of question as asking what happened before the Big Bang, for example, which is literally beyond our comprehension in our current model of reality. The volume of the universe is a theoretically knowable thing, to a certain precision, is it not? There are certainly estimates available. --Mr.98 (talk) 17:13, 21 September 2012 (UTC)[reply]
Note that your link is not to estimates of the volume of the universe, but to the volume of the observable universe. That's completely different; all that can give you is a lower bound. The volume of the entire universe may well be infinite.
That said, Jayron's "unknown unknown" doesn't make sense to me here. So what if we don't know the global topology of the universe? I mean, I think there are only two reasonable choices (it's either R^3 or S^3; there are other models but no explanation of why it would be like that), but even if, by some whim of God, that isn't true, so what? If I ask what is the volume of the ocean, it's not an "unknown unknown" just because we may not know the exact shape of the ocean. --Trovatore (talk) 18:02, 21 September 2012 (UTC)[reply]
Before we can calculate the volume, we'd need to know what shape it is. We don't know that, so it isn't a question of "We know its a big sphere, we just don't know how big of a sphere". That would be a known unknown. If we don't even know the shape of the universe itself, its geometry, we have no means by which to begin starting to calculate the volume. We have different models which presume the universe to be a certain shape, so based on those we could, supposedly, calculate a possible volume, however some of the shapes don't have bounds to them, which makes defining a "volume" rather tricky (of course, with the caveat as discussed in an earlier question that models that deal in those kinds of infinities may have fundemental flaws). So that's why it is an unknown unknown. The observable universe, which you've obliquely cited, is not the universe. Making that assumption is the same as making the assumption that standing on the top of a really big hill near your house and declaring that everything you see is The Whole Earth. --Jayron3217:35, 21 September 2012 (UTC)[reply]
Again, an "unknown unknown" is a very specific thing. It's that which you don't know you don't know. If you know you don't know it, it's not an unknown unknown. The notion of an unknown unknown is a great, useful concept, but not in this sort of thing. It's usually used in trying to extrapolate the future (that is, knowing that there will always be unanticipated problems or ignorances in any complex endeavor), not talk about specific facts. (Logically, anything you can actually say "we don't know" is a known unknown. You can't come up with tangible unknown unknowns, by definition — you don't know what they are!) Anyway, I'm aware that I cited something about the observable universe, but it still means you can get rough estimates on it. It's not that we don't know what questions to ask; we just don't (at the moment) know of what the specific answers are. But there's nothing that makes it clear that we could never figure that out. We can easily say, as Trovatore points out, well, there are a number of possible shapes it could be, and a lot of estimates as to the volume. We don't know which are the right choices to make, so we don't actually know the volume. But these are, again, known unknowns. We know we don't know them. The addition of many known unknowns does not make an unknown unknown. --Mr.98 (talk) 21:44, 22 September 2012 (UTC)[reply]
Can't argue with that. But this ratio seems pretty pointless in that case. A bit like discussing the ratio of neutron charge to universe volume. Ssscienccce (talk) 17:34, 21 September 2012 (UTC)[reply]
Can I ask you a question? Is their any answer you would be satisfied with or do you just want to give us the run around with ambiguous and poorly phased questions? --Aspro (talk) 18:35, 21 September 2012 (UTC)[reply]
165, you'll have to first define a volume for the photon. You seem to have decided that since the rest mass of the photon is zero, that its volume is zero. That doesn't follow from...anything, really. But even if true, so what? To answer your question given your assumptions, yes, the ratio of something that is always 0 to something that is always greater than 0 never changes. It's not a very fascinating revelation. If you're asking if your assumptions are reasonable, maybe someone else can explain to you the absence of a clear definition of volume in quantum mechanics. Someguy1221 (talk) 19:06, 21 September 2012 (UTC)[reply]
Op here, I don't know really. I'm definitely not trying to waste time. I have questions and I ask them to those who I know eventually answer them with some reason. And I try to learn. For some reason I keep getting hung up on how a photon can carry the same energy that it did a million years ago to my eyeball. I can't shake the notion of light traveling "between " space quantum. and black holes being so dense that they become just like the photon existing between the quantum of space.GeeBIGS (talk) 01:39, 22 September 2012 (UTC)[reply]
Comment: 30 or more years ago the rest mass of a photon was one. I know things might have changed a bit since then, but that to my mind destines all my old book into the bin--Aspro (talk) 20:56, 21 September 2012 (UTC)[reply]
Before you do that, maybe you might look through them, and see if you can find this claim that the rest mass of a photon is "one". I suspect that you've misremembered something. If not, it would be interesting to know the context of the claim (and also "one" what?). --Trovatore (talk) 22:12, 21 September 2012 (UTC)[reply]
As I said, it was 30 or more years ago and the batteries in my slide-rule have since gone flat, but here goes: According to Coulomb’s inverse square law the photon may not have a perfectly zero mass ( 10 to the minus 40 or something but not zero). I can't multiply by zero – so call it one. Next. The OP is talking about the whole universe (the largest thing we know of). Therefore, we are not considering just a photon but the total flux or flounce as I think its referred to today. Thus, we have to consider 'invariant mass' and over huge distances. In fact, the very furthest distances we can that we can estimate and the total flux of photons within that huge volume. Please don't ask me to explain that -it was all too long ago. --Aspro (talk) 00:23, 22 September 2012 (UTC)[reply]
As I understand it, the wavelength of a photon increases in direct proportion to the increase in size of the universe. See redshift. But the wavelength of the photon in relation to, say, a molecule hydrogen gas does not change; that molecule will always absorb at the same frequencies. So the spectrum of a distant star - the wavelengths it absorbs at - will also be redshifted, and so you can distinguish absorption that took place there from, say, absorption in some cloud of interstellar gas close to us. Wnt (talk) 21:36, 21 September 2012 (UTC)[reply]
Are you asking if the volume of space per photon is increasing? The answer to that is yes. Most photons are in the cosmic microwave background. The density right now is 412 photons per cubic centimeter (source), or, turning it around, slightly over two cubic millimeters per photon. When the scale factor of the universe was half as large, the density was eight times larger, and the amount of space per photon eight times smaller. That's equally true of protons or electrons, though. And, of course, that's not the amount of space "occupied" by each photon. -- BenRG (talk) 21:51, 21 September 2012 (UTC)[reply]
Hi all, I'm trying to create a simple caption of a photo of a gyroplane, so that even kids can understand. That said, I'm only half sure that *I* understand it. How accurate would this description be? Feel free to edit it, but note that I have a 200 character limit. (It's for a museum exhibit about a county, which happens to include an airport. And note that your edits will appear uncredited, if you make suggestions, because again, 200 character limit.) -- Zanimum (talk) 20:19, 21 September 2012 (UTC)[reply]
"While this gyroplane may look like a helicopter, it isn't. The front motorized rotor provides thrust to hurl it through the air, and the blades on top move without a motor, keeping it up."
A gyroplane is a type of rotorcraft which uses an unpowered rotor in autorotation to develop lift, and an engine-powered propeller, similar to that of a fixed-wing aircraft, to provide thrust.
(EC) I would say "propellor" instead "motorized rotor". Almost every child over the age of 5 knows what an aircraft propellor is. Maybe instead of "keeping it up" you could try "giving lift by acting as a rotating wing" (if that doesn't complicate things too much). I think the concept of the rotor replacing the wings is an important one. Alansplodge (talk) 20:55, 21 September 2012 (UTC)[reply]
(ec) Here's my attempt:
A gyroplane is like a helicopter, but uses an unpowered rotor that
turns by itself, for upward lift, and an engine-powered propeller
in front or back, just like an airplane, to provide forward thrust.
Maybe this is a case of the Americans messing up the English language yet again. Are we talking about a gyroplane or a autogyro. Can you provide a photograph of this mechanical contrivance? Where the propulsion 'air screw' is positioned matters in the rest of the English speaking world as to what we call it. Wikipedia:Systemic bias--Aspro (talk) 21:12, 21 September 2012 (UTC)[reply]
Actually this is the first time I've encountered the word gyroplane. In all the old American movies I've seen, they've always been called autogyros (car-sized Greek food?). Clarityfiend (talk) 00:41, 23 September 2012 (UTC)[reply]
This gyroplane is like a plane with a rotor instead of wings, so that it can fly more slowly without crashing. The engine doesn't turn the rotor, it goes round by itself when the plane moves forward.
Simple description of gyro-plane: A pita folded in the shape of a paper airplane, containing roasted lamb meat, diced tomatoes, lettuce, and onions, drizzled with tzatziki sauce. :-) StuRat (talk) 00:50, 23 September 2012 (UTC)[reply]
I want to know if my solution to a textbook problem has any major problems with it. Here is the problem:
Ethanol has a given density of 0.789 g/mL at 20 degrees Celsius and isopropanol has a given density of 0.785 g/mL at 20 degrees Celsius. A chemist analyzes a substance using a pipet that is accurate to ±0.02 mL and a balance accurate to ±0.003 g. Is this equipment precise enough to distinguish between ethanol and isopropanol?
And here is my solution:
We can calculate with tolerances in the same way we calculate measurements. The mass tolerance of ±0.003 g has three significant figures. The volume tolerance of ±0.02 mL has two significant figures. The density tolerance will therefore have two significant figures. ±0.003 g/±0.02 mL=±0.15 g/mL.In order to distinguish between ethanol and isopropanol, whose densities differ by 0.789 - 0.785 g/ mL, or 0.004 g/ mL, we need a precision smaller than half the difference, or 0.002 g/mL. But we can only measure density to within 0.15 g/mL of the actual value. Therefore, this equipment is not precise enough to distinguish between ethanol and isopropanol.
But what I don't like or feel is right about dividing the tolerances like that is that having a smaller tolerance (more precise) for volume in the denominator blows up (bad) your density tolerance. Shouldn't higher precision (smaller tolerance) of either pipet OR balance result in higher precision (smaller tolerance) of density measurement? Peter Michner (talk) 22:19, 21 September 2012 (UTC)[reply]
See error propagation (especially the section "Example formulas"). Your discomfort is correct. The uncertainty on a ratio is not the same as the ratio of the uncertainties. In fact the uncertainty will depend not only on the tolerance of the scale and pipet but also on how much stuff you plan to measure. If you can measure 100 mL of each then their difference in mass is 0.4 g, an easily measurable amount. So you need to consider the total volume / mass available in order to decide the question. Dragons flight (talk) 22:37, 21 September 2012 (UTC)[reply]
I find the best way to figure the maximum and minimum possible values is to do calculations with whichever values will yield those minimum and maximum values, versus doing math on the tolerances directly. Let's look at a simplified example:
D = m/V
Let's say mass is 1 kg ± 10%, and V = 1 m³ ± 10%. The simple calculation gives D = 1 kg/m³ ± 20%, or 0.8 - 1.2 kg/m³. Now let's do it my way:
D = 0.9/1.1 = .812
D = 1.1/0.9 = 1.222
So, we really have 1.2% less error, on the low side, and over 2.2% more error, in this case, on the high side, using the nominal value as the base. With more terms and higher percentage tolerances, the difference can be even more. Tolerances in exponents and some other mathematical operations are even more critical. StuRat (talk) 23:19, 21 September 2012 (UTC)[reply]
This method overestimates the size of uncertainties for uncorrelated errors. the error calculation should be about D = 1 kg/m³ ± 14%. That's a larger difference than the second order error calculation of 1% or 2% that you pointed out. Dauto (talk) 12:40, 22 September 2012 (UTC)[reply]
I'm not quite sure what you're saying. I've shown that the actual error can be up to 22.2%, in this case. This is not an estimate, this is the actual maximum error. A method which says the error is limited to 14% is therefore incorrect. StuRat (talk) 16:58, 22 September 2012 (UTC)[reply]
An experimental value quoted as x±y does not mean "between x−y and x+y" but rather "a mean of x and a standard deviation of y". (Sometimes y is two standard deviations since that corresponds to a confidence level of about 95%.) This is a compact way of representing a probability distribution that's often approximately a Gaussian (bell curve) centered on x. Dauto's answer is correct in that situation. Your answer is correct under the very different assumptions you made. -- BenRG (talk) 19:12, 22 September 2012 (UTC)[reply]
You can't always assume a bell-shaped distribution. For example, resistors with a ±20% tolerance are likely to be either 10-20% high or low, and not within 10% of the nominal value, since they test them all and sell those found within the ±10% range separately, for more money. StuRat (talk) 20:52, 22 September 2012 (UTC)[reply]
I asked the same question on Physics Stackexchange and got this answer, which concludes that given the premise that x±y means the value is between x-y and x+y, there is a 98% chance of being able to tell ethanol from isopropanol given the tolerances of the equipment. That is the interpretation of x±y given in the particular textbook I'm looking at (chapter 1, problem 71), and I have never heard of the interpretation of that notation meaning a bell curve distribution. Of course, I don't live in the world of science. I think the answerer of my question there accidentally used 0.002 g instead of 0.003 g for the balance tolerance, but that is a different issue. Peter Michner (talk) 21:43, 22 September 2012 (UTC)[reply]
That may indeed be the interpretation expected by the textbook but that's not how experimental uncertainties work in practice. The interpretation as described by BenRG is the one that is actually used by experimental physicists because it gives more reliable results in actual experiments. That's the interpretation I used to calculate the 14% compounded uncertainty in the example above. Dauto (talk) 04:26, 23 September 2012 (UTC)[reply]
Edit: I tried to give just as much information as is relevant to my problem, but I guess one detail from part (a) of the textbook problem (what I described was part (b)), which said the nominal sample volume was 15.00 mL, must carry over to part (b). So I think I assume the nominal volume of my sample is 15.00 mL plus or minus 0.02 mL. Peter Michner (talk) 01:24, 22 September 2012 (UTC)[reply]
Watch out with a term like "significant figures", you can say your result has 3 4 or 5 significant figures, you don't say that about a tolerance by itself. May lose you points on an exam. Ssscienccce (talk) 09:37, 22 September 2012 (UTC)[reply]
Like, with one minor change: "rotor" -> "overhead rotor". Clarityfiend (talk) 00:43, 23 September 2012 (UTC)